Author(s): HAN Haiying
Pages: 6-8
Year: 2006 Issue:  3
Journal: School Journal of Mathematics
Keyword:  求值; 条件分式; 条件式; 已知量; 变形; 例析; 已知条件; 练习; 等比定理; 技巧;
Abstract: <正>一、比值法如果已知的条件式中出现连比的形式,可通过设其比值为参数k, 建立分子和分母的关系式,然后经过适当的变形,求出分式的值。例1 己知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz≠0,求(x+y)(z+x)/xyz 的值．解：设x+y-z/z=x-y+z=-x+y+z=k,则 x+y-z=kz,x-y+z=ky,-x+y+z=kx．上面三式相加,得x+y+z=k(x+y+z)．① (1)当x+y+z=0时,x+y=-z,y+z=-x,z+x=-y．∴(x+y)(y+z)(z+x)/xyz=(-z)(-x)(-y)/xyz=-1．